Check if a function is a method of a class?

Question

Is there a way to determine if a function is a method of a certain class?

I have a class A with a method doesMethodBelongHere, that takes a function as an argument method. I want to determine that method is an actual method of A.

class A {
  methodA() {
    console.log('method of A');
  }
  
  doesMethodBelongHere(method) {
    // it should return true if `method` argument is a method of A
    return Object.values(this).includes(method);
  }
}

const a = new A(); 
console.log(a.doesMethodBelongHere(a.methodA)); // should return true

Possible duplicate of [How to check if function exists in JavaScript?](https://stackoverflow.com/questions/1042138/how-to-check-if-function-exists-in-javascript) — Liam, Oct 19 '18 at 14:35
@Liam Linked duplicate asks the general question, "Does this method exist _at all_?" - The question OP is asking is, "How do I confirm A is a method of class B?". — Lewis, Oct 19 '18 at 14:44
It should return true if it's specifically a method of the specific instance of A? or generally a method of class A? — Reuven Chacha, Oct 19 '18 at 14:53

Ori Drori · Accepted Answer · 2018-11-01T23:26:31.273

You can use Object.getPrototypeOf() to get the prototype. Then iterate the prototype properties using for...of, and Object.getOwnPropertyNames(). If the method is equal to one of the methods on the prototype return true:

class A {
  methodA() {
    console.log('method of A');
  }

  doesMethodBelongHere(method) {
    // get the prototype
    const proto = Object.getPrototypeOf(this);
    
    // iterate the prototype properties, and if one them is equal to the method's reference, return true
    for(const m of Object.getOwnPropertyNames(proto)) {
      const prop = proto[m];
      if(typeof(prop) === 'function' && prop === method) return true;
    }
    
    return false;
  }
}

const a = new A();
Object.assign(a, { anotherMethod() {} }); 
a.anotherMethod2 = () => {};

console.log(a.doesMethodBelongHere(a.methodA)); // should return true

console.log(a.doesMethodBelongHere(a.anotherMethod)); // should return false

console.log(a.doesMethodBelongHere(a.anotherMethod2)); // should return false

Extended classes:

This solution will also handle methods that comes from an extended class:

class A {
  methodA() {
    console.log('method of A');
  }

  doesMethodBelongHere(method) {
    let proto = this;
    
    // iterate the prototypes chain
    while (proto = Object.getPrototypeOf(proto), proto && proto !== Object) {
      // iterate the prototype properties, and if one them is equal to the method's reference, return true
      for (const m of Object.getOwnPropertyNames(proto)) {
        const prop = proto[m];
        if (typeof(prop) === 'function' && prop === method) return true;
      }
    }

    return false;
  }
}

class B extends A {}

class C extends B {}

const c = new C();
Object.assign(c, {
  anotherMethod() {}
});

c.anotherMethod2 = () => {};

console.log(c.doesMethodBelongHere(c.methodA)); // should return true

console.log(c.doesMethodBelongHere(c.anotherMethod)); // should return false

console.log(c.doesMethodBelongHere(c.anotherMethod2)); // should return false

score 0 · Answer 2 · answered Oct 19 '18 at 14:40

0

    class A {
      constructor() {
        this.methodA = this.methodA.bind(this);
        this.doesMethodBelongHere = this.doesMethodBelongHere.bind(this);
      }
     methodA() {
        console.log('method of A');
      }
      
      doesMethodBelongHere(method) {
        // it should return true if `method` argument is a method of A
        return Object.values(this).includes(method);
      }
    }

    const a = new A(); 
    console.log(a.doesMethodBelongHere(a.methodA)); // should return true

This was not bound to your class in your doesMethodBelongHere.

answered Oct 19 '18 at 14:40

Matthi

1,073
8
19

Can you elaborate on that? Can it be achieved without explicit binding? – Serhii Holinei Oct 19 '18 at 14:41
`this` shifted in the moment of instanciating. Binding `this` to your class keeps the context at the right place. – Matthi Oct 19 '18 at 14:51

Rajib Chy · Answer 3 · 2018-10-19T14:57:42.807

0

You can use typeof operator

let isfn = "function" === typeof ( a.methodA );//isfn should be true
isfn = "function" === typeof ( a["methodA"] );//isfn should be true
isfn = "function" === typeof ( a["methodAX"] );//isfn should be false

Edit

doesMethodBelongHere( method ) {
    return  "function" === typeof ( this[method.name] )
}

edited Oct 19 '18 at 14:57

answered Oct 19 '18 at 14:54

Rajib Chy

800
10
22

But I pass a reference to a function, the trick is to get its name. – Serhii Holinei Oct 19 '18 at 14:55

Reuven Chacha · Answer 4 · 2018-10-19T19:46:48.183

I'd suggest to use the following implementation:

Use Object.getOwnPropertyNames on the constructor prototype (the same accessing A.prototype, but in a more generic approach) in order to iterate class methods.
Get method name using method.name

Using Array.some, find whether (1) includes the name of the given method.

class A {

    constructor() {
        this.a = 2;
        this.bb = 3;
    }
    methodA() {
        console.log('method of A');
    }

    doesMethodBelongHere(method) {
        // it should return true if `method` argument is a method of A
        return this.constructor.prototype[method.name] === method;
    }
}

Method names equivalence doesn't imply the same methods. – Serhii Holinei Oct 19 '18 at 15:28 — Serhii Holinei, Oct 19 '18 at 15:28

Check if a function is a method of a class?

4 Answers4

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